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1.26 kB
| #Greg Considine | |
| #Project Euler -- Problem 9 | |
| import math | |
| def coprime(x,y): | |
| min = x if x < y else y | |
| for i in range(2, math.ceil(math.sqrt(min)) + 1): | |
| if x % i == 0 and y % i == 0: | |
| return False | |
| return True | |
| def odd(x): | |
| if x % 2 == 0: | |
| return False | |
| else: | |
| return True | |
| def pythagorean(x,y,z): | |
| if (math.pow(x,2) + math.pow(y,2)) == math.pow(z,2): | |
| return True | |
| else: | |
| return False | |
| m = 2 | |
| n = 1 | |
| a = b = c = 0 | |
| total = 0 | |
| found = False | |
| while found == False: | |
| while n < 100 and found == False: | |
| if coprime(m,n) and odd(m-n): | |
| a = (math.pow(m,2) - math.pow(n,2)) | |
| b = 2 * m * n | |
| c = (math.pow(m,2) + math.pow(n,2)) | |
| if pythagorean(a,b,c): | |
| total = a + b + c | |
| if total == 1000: | |
| found = True | |
| print(a * b * c) | |
| n += 1 | |
| else: | |
| n += 1 | |
| m += 1 | |
| n = 1 | |
| ''' | |
| This was probably the most challenging problem yet and my solution isn't very | |
| elegant. I had to read up on Pythagorean triples and coprime numbers. I used | |
| Euclid's formula to generate pythagorean triples with incrementing values of | |
| m and n. I was going to manually increase the limits of n if no solution | |
| was found -- not exactly a one-size-fits-all algorithm but it works. | |
| ''' | |